Three blocks of masses `m_(1), m_(2)` and `m_(3)` kg are placed in contact with each other on a frictionless table. A force F is applied on the heaviest mass `m_(1)` , the acceleration of `m_(3)` will be `:`

Options

A. `(m_(2))/(m_(1)+m_(2)+m_(3))T`
B. `(m_(3))/(m_(1)+m_(2)+m_(3))T`
C. `(m_(1)+m_(2))/(m_(1)+m_(2)+m_(3))T`
D. `(m_(2)+m_(3))/(m_(1)+m_(2)+m_(3))T`

Answer

C

Explanation

`T=(m_(1)+m_(2)+m_(3))a`
`T-T_(1)=m_(3)a`
`T_(1)=T-m_(3)a`
`=T-(m_(3))/((m_(1)+m_(2)+m_(3))xxT)=(T(m_(1)+m_(2)))/((m_(1)+m_(2)+m_(3)))`